In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Asked by Abhisek | 1 year ago | 183

Given

The work done (W) on the system while the gas changes from state A to state B is 22.3 J

This is an adiabatic process.

Therefore, change in heat is zero.

So,

ΔQ = 0

ΔW = – 22.3 (Since the work is done on the system)

From first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where,

ΔU = Change in the internal energy of the gas

Hence,

ΔU = ΔQ – ΔW

= 0 – (-22.3 J)

We get,

ΔU = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

ΔQ = 9.35 cal

= 9.35 x 4.19

On calculation, we get,

= 39.1765 J

Heat absorbed, ΔQ = ΔU + ΔW

Thus,

ΔW = ΔQ – ΔU

= 39.1765 – 22.3

We get,

= 16.8765 J

Hence, 16.88 J of work is done by the system

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